Here's another counter-example, taken from Loop Groups (p 128).
Consider the space of continuous functions on the circle and define
$$
f(\theta) = \sum_{k \gt 1} \frac{\sin k \theta}{k \log k}
$$
The positive part of this function is
$$
f_+(\theta) = \frac{1}{2 i} \sum_{k \gt 1} \frac{e^{i k \theta}}{k \log k}
$$
which is unbounded near $\theta = 0$.
Let's move on to the other part of your question: when can we ensure that the splitting exists? What you are asking for is that $V$ be the direct sum of two subspaces, $V_-$ and $V_+$. Of course, if you start with two abstract vector spaces, $V_-$ and $V_+$, both of which admit symmetric, positive definite bilinear forms, say $b_-$ and $b_+$ respectively, then you can construct an example by taking $V = V_- \oplus V_+$ and taking $-b_- + b_+$. This shows that you can get this situation to work with quite awful spaces, but the point is that all the awfulness of $V$ divides nicely into awfulness of $V_-$ plus awfulness of $V_+$.
Presumably, though, you are more interested in the case where you start with $V$ and the quadratic form. Maybe this quadratic form can be fairly arbitrary (perhaps varies in some space of quadratic forms). In this situation, you would want conditions on $V$ that guarantee that the splitting occurs without too much fuss.
Let's examine the question from the other end: suppose that $V = V_- \oplus V_+$. Then by changing the sign of the form on $V_-$, we obtain a positive-definite symmetric bilinear form on $V$. This usually goes by the name of an inner product as we're over $\mathbb{R}$. So the problem reduces to finding complements of subspaces in inner product spaces. To guarantee this, you want completeness. Then your bilinear form is related to the original inner product by the operator $2P_+ - I$ where $P_+$ is the orthogonal projection on to $V_+$.
So what you want is to be working with a Hilbert space and the space of self-adjoint square-roots of the identity.
As I said, this isn't an "if and only if". But it is a simple condition that quite often holds. It can be further relaxed since it's enough that the inner product induced by the bilinear form and the original inner product be merely equivalent rather than equal, but I'll leave those details as an exercise.
Edit: From your other question related to this it seems as though you are particularly interested in the case where one of the factors is finite dimensional. In that case, the splitting always holds.
Here's a proof that $\exp(z)$ is not a characteristic function using the product expansion for the determinant, which is essentially equivalent to Lidskii's theorem stating that the trace of a trace class operator is the sum of its eigenvalues.
If $T$ is a trace class operator on a Hilbert space $\mathcal{H}$ with eigenvalues $\lambda_n$ (counted up to algebraic multiplicity), then $\sum_n\vert\lambda_n\vert < \infty$ and
$$
{\rm det}(1+zT)\equiv\sum_{k=0}^\infty{\rm Tr}(\Lambda^kT)z^k=\prod_n(1+z\lambda_n)
$$
for all $z\in\mathbb{C}$.
See Trace Ideals and Their Applications, Theorem 3.7. Actually, they use this result to prove Lidskii's theorem, that ${\rm Tr}(T)=\sum_n\lambda_n$.
Then, ${\rm det}(1+zT)$ cannot be equal to $\exp(z)$ everywhere. This is because $\exp$ has no zeros, so $\lambda_n$ would have to be all zero, giving ${\rm det}(1+zT)=1$.
Best Answer
A simple proof:
Assume $\mathcal{H}=\mathcal{H}_+\oplus\mathcal{H}_-$, with $( A u| u)>0$ (resp. $<0$) for all nonzero $u\in\mathcal{H}_+$ (resp. $u\in\mathcal{H}_-$). We want to show $\ker A=0$. To this end, assume $Au=0$ and write $u=u_++u_-$ with $u_\pm\in\mathcal{H}_\pm$. If either $u_+$ or $u_-$ is zero, the assumption easily implies $u=0$. Otherwise, we have $$ 0=(Au|u_+)=(Au_+|u_+)+(Au_-|u_+)>(Au_-|u_+), $$ $$ 0=(Au|u_-)=(Au_+|u_-)+(Au_-|u_-)<(Au_+|u_-). $$ In view of the self-adjointness, we get a contradiction.